Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int const STEP = 241;
int n, h;
double getTrapezeArea(double d, double t, double h) {
  return (d + t) * h / 2.0;
}
void solve() {
  scanf("%d %d", &n, &h);
  double S = h * 1.0 / 2;
  double needS = S / n;
  double d = 1.0;
  double curH = 0;
  vector<double> ans;
  for (int i = 0; i < n - 1; ++i) {
    double l = curH;
    double r = h;
    for (int j = 0; j < STEP; ++j) {
      double m = (l + r) / 2;
      if (getTrapezeArea(d, (h - m) / h, m - curH) >= needS)
        r = m;
      else
        l = m;
    }
    ans.push_back(h * 1.0 - r);
    d = (h - r) / h;
    curH = r;
  }
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < (int)(ans.size()); ++i) printf("%.15lf ", ans[i]);
}
int main() {
  solve();
  return 0;
}